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insens350 [35]
3 years ago
8

Can someone help me please

Chemistry
1 answer:
MArishka [77]3 years ago
3 0

Answer:

Group 4A (or IVA) of the periodic table includes the nonmetal carbon (C), the metalloids silicon (Si) and germanium (Ge), the metals tin (Sn) and lead (Pb), and the yet-unnamed artificially-produced element ununquadium (Uuq).

The Group 4A elements have four valence electrons in their highest-energy orbitals (ns2np2). Carbon and silicon can form ionic compounds by gaining four electrons, forming the carbide anion (C4-) and silicide anion (Si4-), but they more frequently form compounds through covalent bonding. Tin and lead can lose either their outermost p electrons to form 2+ charges (Sn2+, the stannous ion, and Pb2+, the plumbous ion) or their outermost s and p electrons to form 4+ charges (Sn4+, the stannic ion, and Pb4+, the plumbic ion).

Carbon (C, Z=6).

Carbon is most familiar as a black solid is graphite, coal, and charcoal, or as the hard, crystalline diamond form. The name is derived from the Latin word for charcoal, carbo. It is found in the Earth's crust at a concentration of 480 ppm, making it the 15th most abundant element. It is found in form of calcium carbonate, CaCO3, in minerals such as limestone, marble, and dolomite (a mixture of calcium and

Explanation:

<em><u>T</u></em><em><u>H</u></em><em><u>I</u></em><em><u>S</u></em><em><u> </u></em><em><u>A</u></em><em><u>L</u></em><em><u>L</u></em><em><u> </u></em><em><u>I</u></em><em><u> </u></em><em><u>K</u></em><em><u>N</u></em><em><u>O</u></em><em><u>W</u></em>

<u>E</u><u>N</u><u>J</u><u>O</u><u>Y</u><u> </u><u>THE</u><em><u> </u></em><em><u>A</u></em><em><u>N</u></em><em><u>S</u></em><em><u>W</u></em><em><u>E</u></em><em><u>R</u></em>

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Find the number of moles of barium iodide if you have 5.23 x 1024 formula units of Bal2
Oksi-84 [34.3K]

Answer:

8.68 moles of BaI₂

Explanation:

Given data:

Number of moles of BaI₂ = ?

Number of formula units = 5.23× 10²⁴

Solution:

By using Avogadro number,

1 mole of any  substance contain 6.022× 10²³ formula units.

5.23× 10²⁴ formula units of BaI₂ × 1 mol / 6.022× 10²³ formula units

0.868 × 10¹ moles of BaI₂

8.68 moles of BaI₂

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A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution
Marianna [84]

Answer:

0.136g

Explanation:

A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?

Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)

Initial mole of Co(NO3)2  =\frac{mass}{molar mass}

=\frac{5.00}{182.94} \\\\=0.02733mol

Mole of Co(NO3)2 in final solution

=\frac{4.00}{100}\times 0.02733\\\\=0.04\times 0.02733\\\\= 0.001093mol

Mole of  NO3- in final solution = 2 x Mole of Co(NO3)2

=2\times 0.001093\\\\=0.002186mol

Mass of  NO3- in final solution is mole x Molar mass of NO3

=0.002186\times62.01\\\\=0.136g

6 0
3 years ago
A mixture of noble gases [helium (MW 4), argon (MW 40), krypton (MW 83.8), and xenon (MW 131.3)] is at a total pressure of 150 k
kodGreya [7K]

Answer:

a) 1,6%

b) 64,775 g/mol

c) 3,6×10⁻² M

d) 2,3×10⁻³ g/mL

Explanation:

a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:

  • Helium: 0,25 moles ×\frac{4 g}{1 mol} = 1 g of Helium
  • Argon: 0,25 moles ×\frac{40 g}{1 mol} = 10 g of Argon
  • Krypton: 0,25 moles ×\frac{83,8 g}{1 mol} = 20,95 g of krypton
  • Xenon: 0,25 moles ×\frac{131,3 g}{1 mol} = 32,825 g of Xenon

Total grams: 1g+10g+20,85g+30,825g= 62,675 g

Mass fraction of helium: \frac{1 gHelium}{62,675 g} × 100 = <em>1,6%</em>

<em />

<em>The mass fraction of Helium is 1,6%</em>

<em />

<em>b)</em><em>  </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:

\frac{4+40+83,8+131,3}{4} = 64,775 g/mol

c) The molar concentration is possible to know ussing ideal gas law, thus:

\frac{P}{R.T} = M

Where:

P is pressure: 150 kPa

R is gas constant: 8,3145\frac{L.kPa}{K.mol}

T is temperature: 500 K

And M is molar concentration. Replacing:

M = 3,6×10⁻² M

d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:

3,6×10⁻² \frac{mol}{L} × \frac{64,775 g}{mol} × \frac{1L}{1000 mL} =

2,3×10⁻³ g/mL

I hope it helps!

7 0
3 years ago
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